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\(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(P\left(x\right)=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(P\left(x\right)=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(P\left(x\right)=x-\sqrt{x}-2\sqrt{x}-2+2\sqrt{x}+2\)
\(P\left(x\right)=x-\sqrt{x}\)
Ta có : \(\dfrac{P\left(x\right)}{2020\sqrt{x}}=\dfrac{x-\sqrt{x}}{2020\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2020\sqrt{x}}=\dfrac{\sqrt{x}-1}{2020}\)
Để \(\dfrac{P\left(x\right)}{2020\sqrt{x}}min\Leftrightarrow\dfrac{\sqrt{x}-1}{2020}min\Leftrightarrow\sqrt{x}-1\) min (vì 2020 > 0)
Lại có : \(\sqrt{x}-1\ge-1\forall x\)
Dấu "=" xảy ra <=> x = 0
Vậy Min\(\dfrac{P\left(x\right)}{2020\sqrt{x}}=\dfrac{-1}{2020}\Leftrightarrow x=0\)
Ta có : \(\sqrt{x+1}\) có nghĩa khi `x >= -1` Từ đk ta có :
\(x+2\left(1+\sqrt{x+1}\right)=x+1+2\sqrt{x+1}+1=\left(\sqrt{x+1}+1\right)^2\)
\(\Leftrightarrow\sqrt{x+2\left(1+\sqrt{x+1}\right)}=\sqrt{x+1}+1\)
\(x+2\left(1-\sqrt{x+1}\right)=x+1-2\sqrt{x+1}+1=\left(\sqrt{x+1}-1\right)^2\\ \Leftrightarrow\sqrt{x+2\left(1-\sqrt{x+1}\right)}=\left|\sqrt{x+1}-1\right|\)
Ta có : \(y=\sqrt{x+1}+1+\left|\sqrt{x+1}-1\right|\) `(1)`
Ta bỏ dấu \(\left|\right|\) ở `1`
Ta có TH :
`-1<= x <= 0` ; lúc này \(\sqrt{x+1}-1\le0\)
nên : \(\left|\sqrt{x+1}-4\right|=1-\sqrt{x+1}\)
`1` trở thành : `y=2`
`x>0` lúc này \(\sqrt{x+1}-1>0\) nên
\(\left|\sqrt{x+1}-1\right|=\sqrt{x+1}-1\)
`1` trở thành : \(y=2\sqrt{x+1}>2\left(x>0\right)\)
Vì : \(y=\left\{{}\begin{matrix}2khi-1\le x\le0\\2\sqrt{x+1}kh\text{i}>0\end{matrix}\right.\)
gtnn của `y=2` với mọi \(x\in\left[-1;0\right]\)
ĐKXĐ: \(x\ge0;x\ne1\)
\(M=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(M=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\ge1-\dfrac{2}{0+1}=-1\)
\(M_{min}=-1\) khi \(x=0\)
I don't now
sorry
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C=|x-2021|+|1-x|>=|x-2021+1-x|=2020
Dấu = xảy ra khi 1<=x<=2021
1:
a: \(A=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
căn x+1>=1
=>2/căn x+1<=2
=>-2/căn x+1>=-2
=>A>=-2+1=-1
Dấu = xảy ra khi x=0
b: 
\(A=\)\(\sqrt{x+2\left(1+\sqrt{x+1}\right)}+\sqrt{x+2\left(1-\sqrt{x+1}\right)}\) (đk: \(x\ge-1\))
\(=\sqrt{\left(x+1\right)+2\sqrt{x+1}+1}+\sqrt{\left(x+1\right)-2\sqrt{x+1}+1}\)
\(=\sqrt{\left(\sqrt{x+1}+1\right)^2}+\sqrt{\left(\sqrt{x+1}-1\right)^2}\)
\(=\sqrt{x+1}+1+\left|\sqrt{x+1}-1\right|\)
\(=\left[{}\begin{matrix}\sqrt{x+1}+1+\sqrt{x+1}-1;\sqrt{x+1}\ge1\\\sqrt{x+1}+1-\left(\sqrt{x+1}-1\right);\sqrt{x+1}< 1\end{matrix}\right.\)
\(=\left[{}\begin{matrix}2\sqrt{x+1};x\ge0\\2;-1\le x< 0\end{matrix}\right.\)
Có \(2\sqrt{x+1}\ge2\) tại \(x\ge0\)
\(\Rightarrow\min\limits_{x\ge0}A=2\)
Dấu = xảy ra <=> x=0 mà tại \(-1\le x< 0\) thì A=2
Vậy giá trị nhỏ nhất của biểu thức là 2 tại x=0 hoặc \(-1\le x< 0\)
(Ủa đề zì kì)
\(ĐKXĐ:x\ge-1\)
Đặt \(A=\sqrt{x+2\left(1+\sqrt{x+1}\right)}+\sqrt{x+2\left(1-\sqrt{x+1}\right)}\)
\(=\sqrt{x+1+2\sqrt{x+1}+1}+\sqrt{x+1-2\sqrt{x+1}+1}\)
\(=\sqrt{\left(\sqrt{x+1}+1\right)^2}+\sqrt{\left(\sqrt{x+1}-1\right)^2}\)
\(=\left|\sqrt{x+1}+1\right|+\left|\sqrt{x+1}-1\right|\)
\(=\left|\sqrt{x+1}+1\right|+\left|1-\sqrt{x+1}\right|\)
\(\ge\left|\sqrt{x+1}+1+1-\sqrt{x+1}\right|=2\)
Dấu "=" xảy ra khi \(\left(\sqrt{x+1}+1\right)\left(1-\sqrt{x+1}\right)\ge0\)
\(\Leftrightarrow1-\sqrt{x+1}\ge0\)
\(\Leftrightarrow\sqrt{x+1}\le1\)
\(\Leftrightarrow x\le0\). Mà \(x\ge-1\) Nên \(-1\le x\le0\)
Vậy Min \(A=2\) khi \(-1\le x\le0\)